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One-Answer Puzzle Theory Question Thread

I am trying to learn 2GR and am having trouble understanding how slice moves (and wide moves, by extension) affect EO. For ZZ I learned the definition that an edge is good if it is in the <R, U, L, D, F2, B2> group. Unfortunately, following an M, E, or S move, there are two ways in which a solved edge can be defined. The first definition is that an edge is solved iff it is adjacent to both centers of its component colors and one of its colors is adjacent to the same colored center, and the second definition is that an edge is solved iff it would satisfy the previous definition assuming that all center pieces are in WCA orientation.

Here are the two metrics on a solved cube in WCA orientation:

Relative to Centers Definition:
M': All edges in the M slice are still solved, and therefore can be solved in <R, U, L, D, F2, B2>, and the remaining eight edges can be trivially solved in 2 FTM in <R, L>, and therefore can be solved in <R, U, L, D, F2, B2>. Thus M' does not affect EO. Therefore (by induction) <M> does not affect EO.

E': All edges in the E slice are still solved, and therefore can be solved in <R, U, L, D, F2, B2>, and the remaining eight edges can be trivially solved in 2 FTM in <U, D>, and therefore can be solved in <R, U, L, D, F2, B2>. Thus E' does not affect EO. Therefore (by induction) <E> does not affect EO.

S: All edges in the S slice are still solved, and therefore can be solved in <R, U, L, D, F2, B2>, but the remaining eight edges cannot be solved in <R, U, L, D, F2, B2>. Thus S flips the orientation of the F and B edges.


Relative to Cube Orientation Definition:
M': All edges in the M slice are now bad because performing F2 B2 leaves the four M edges permuted correctly, but oriented incorrectly. The remaining eight edges are already in their solved position and therefore can be solved in <R, U, L, D, F2, B2>. Thus M' flips the orientation of the M edges.

E': All edges in the E slice are now bad because performing F2 B2 leaves the four E edges permuted correctly, but oriented incorrectly. The remaining eight edges are already in their solved position and therefore can be solved in <R, U, L, D, F2, B2>. Thus E' flips the orientation of the E edges.

S: All edges in the S slice are now bad because performing L2, U2, R2, xor D2 leaves the Yellow-Orange, Orange-White, Red-White, and Red-Yellow E edges respectively permuted correctly, but oriented incorrectly. The remaining eight edges are already in their solved position and therefore can be solved in <R, U, L, D, F2, B2>. Thus S flips the orientation of the S edges.


  • So an r move, would either leave EO the same, or change the orientation of the M-slice edges.
  • A u move would either leave EO the same, or change the orientation of the E-slice edges.
  • An f move would either maintain the orientation of the F-face edges (F or F' changes F-face EO, but S according to the Relative Centers definition would also change F-face EO, so it cancels out) while flipping the orientation of B-face edges and maintaining the orientation of the S-slice edges, or change the orientation of the F-face and S-slice edges.

So which is it, and how do you keep track of it in a solve? I've been going through some of the example solves in the 2GR doc, and the EO-Pair solutions seem incredibly arcane to me; even more-so than EO-Line, which I can at least follow even though I can't plan it in inspection of find an optimal solution pretty much ever on my own. Why do the solutions rely so much on <Rr, Uu, Ff> to solve EO-pair? Do any ZZ solves do EO-Line like that? How does one begin to understand how teoidus (the creator of 2GR) up with his EO-Pair solutions? Even if he used a computer solver, I still think it's reasonable to assume he wouldn't have used those solutions to EO-pair in his example solutions if he didn't think that they were reasonable examples of how to solve for a human. So while a human solution might take a few more moves with only 15 seconds to plan, it should still use some wide moves most of the time based on the examples provided.
 
In ZZ, the first step is to orient edges so that the cube is the <R, U, L, D, F2, B2> group. Is this the same as the <R, U, L, D> group?

In other words, is it possible to solve a cube with the scramble F2 using only <R, U, L, D> ?

Edit: After typing this I tried it with brute force and confirmed that it is possible, just by doing a regular ZZ solve. Follow-up question: What is the optimal solution to solving F2 with <R, U, L, D>?
 
MethodNeutral said:
In ZZ, the first step is to orient edges so that the cube is the <R, U, L, D, F2, B2> group. Is this the same as the <R, U, L, D> group?

In other words, is it possible to solve a cube with the scramble F2 using only <R, U, L, D> ?

Edit: After typing this I tried it with brute force and confirmed that it is possible, just by doing a regular ZZ solve. Follow-up question: What is the optimal solution to solving F2 with <R, U, L, D>?
Click to expand...
Here are all of the move-optimal solutions:
U L' U R U R' U2 L2 D' L D L U' L (14f*)
U' R U' L' U' L U2 R2 D R' D' R' U R' (14f*)
R U' R D R D' R2 U2 L' U L U R' U (14f*)
R' D R' U' R' U R2 D2 L D' L' D' R D' (14f*)
D R' D L D L' D2 R2 U' R U R D' R (14f*)
D' L D' R' D' R D2 L2 U L' U' L' D L' (14f*)
L D' L U L U' L2 D2 R' D R D L' D (14f*)
L' U L' D' L' D L2 U2 R U' R' U' L U' (14f*)
 
MethodNeutral said:
In ZZ, the first step is to orient edges so that the cube is the <R, U, L, D, F2, B2> group. Is this the same as the <R, U, L, D> group?

In other words, is it possible to solve a cube with the scramble F2 using only <R, U, L, D> ?

Edit: After typing this I tried it with brute force and confirmed that it is possible, just by doing a regular ZZ solve. Follow-up question: What is the optimal solution to solving F2 with <R, U, L, D>?
Click to expand...

I know this has already been answered, and this isn't the solution asked for, but this reminded me of Chris Hardwick's signature: R L F2 B2 L' R' U R L B2 F2 L' R'

It blew my mind when I first tried it out.
 
Hello, math related question! I'm working on a new LS method for 3x3 where the corners are oriented and permuted, kind of like OLS except instead of solving the last pair + OLL, it solves the last pair + corners(for CFCE instead of CFOP). But I was wondering, how many cases are there? I'm mostly wondering for R U R'(opposite) and R U' R'(paired) inserts, so more like VLS and HLS combined except for corners.

Thanks!
 
ProStar said:
Hello, math related question! I'm working on a new LS method for 3x3 where the corners are oriented and permuted, kind of like OLS except instead of solving the last pair + OLL, it solves the last pair + corners(for CFCE instead of CFOP). But I was wondering, how many cases are there? I'm mostly wondering for R U R'(opposite) and R U' R'(paired) inserts, so more like VLS and HLS combined except for corners.

Thanks!
Click to expand...
I don’t remember the number off the top of my head but you’re not the first to ask about this. It is quite a lot of algorithms, which is why it hasn’t really been developed. Another consequence of this is that because there are so many algs in the set, the quality of the algs is usually pretty poor.
 
OreKehStrah said:
I don’t remember the number off the top of my head but you’re not the first to ask about this. It is quite a lot of algorithms, which is why it hasn’t really been developed. Another consequence of this is that because there are so many algs in the set, the quality of the algs is usually pretty poor.
Click to expand...

It could be something like OLS though, where people only learn part of it, not all of it.
 
ProStar said:
Hello, math related question! I'm working on a new LS method for 3x3 where the corners are oriented and permuted, kind of like OLS except instead of solving the last pair + OLL, it solves the last pair + corners(for CFCE instead of CFOP). But I was wondering, how many cases are there? I'm mostly wondering for R U R'(opposite) and R U' R'(paired) inserts, so more like VLS and HLS combined except for corners.

Thanks!
Click to expand...

162 each, for a total of 324.
This is a thing in roux, although it has affects M slice algs. It is not known at all, because it is recently posted on reddit and no one took notice really. I find it surprising that it is a thing in Roux but not CFOP, even though it is much for relevant in a roux solve.
 
In the Wiki database it is stated that there are 21 PLL manuevers. Doing the math, there are 4! * 4! / 2 = 288 OLL cubes which may be reduced at most 4 fold by C4 symmetry. I get 83 non-trivial cubes after symmetry reduction. How do you get only 21 maneuvers?
 
How exactly did you end up with 83? The simplest reduction is dividing by 4 to account for the 4 possible AUFs for each case where OLL is solved, which already reduces to 72. (Since doing, say, an H-perm followed by any U move still results in an H-perm.) Now symmetry can be accounted for. Note that for something fully symmetrical like an H-perm, you can do any number of y rotations prior to the H-perm and still end up in the same state. On the other hand, for something like a T-perm which has no symmetry, executing from different angles results in different cases, all of which appear in our current 72. That is to say, the T-perm case is counted four times when it should only be counted once.

Different PLLs have different symmetry, so some will be counted only once (like an H-perm), some will be counted twice (like an E-perm), and many will be counted four times. Once these duplicates are subtracted, it should come out to 21.

(I kinda wrote this up quickly so I haven't done a rigorous proof to end up at 21, but I know from experience that there are only 21 PLLs and this reasoning makes logical sense to me. Someone please correct me if anything here is wrong!)
 
DGCubes said:
Different PLLs have different symmetry, so some will be counted only once (like an H-perm), some will be counted twice (like an E-perm), and many will be counted four times. Once these duplicates are subtracted, it should come out to 21.

(I kinda wrote this up quickly so I haven't done a rigorous proof to end up at 21, but I know from experience that there are only 21 PLLs and this reasoning makes logical sense to me. Someone please correct me if anything here is wrong!)
Click to expand...
22 if you include the solved case.

Bruce MacKenzie said:
In the Wiki database it is stated that there are 21 PLL manuevers. Doing the math, there are 4! * 4! / 2 = 288 OLL cubes which may be reduced at most 4 fold by C4 symmetry. I get 83 non-trivial cubes after symmetry reduction. How do you get only 21 maneuvers?
Click to expand...
To restate what DG said a bit more formally, PLL cases are considered to be identical if you can compose with U moves to get from one to the other, i.e. \( X \sim Y \) iff \( X = \mathrm U^a\circ Y\circ \mathrm U^b \) for some \( a, b\in\{0,1,2,3\} \). Thus there should be sixteen symmetries, not just the four rotational symmetries (where you restrict it to \( a+b=0 \)).
 
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